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2 years ago

Can somebody help me with this problem whoever helps me first gets a thanks ❤️

Mathematics

2 answers:

marysya [2.9K]2 years ago

7 0

Step-by-step explanation:

top left has four right angles

kow [346]2 years ago

6 0

It’s the first one. Rectangle

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I want the answer to this , zoom in if you can't see well

galben [10]

We are going to need to know the formula for the lateral surface area and volume of a cylinder in order to find our answers. Below are the equations:

$A_L = 2 \pi r h$

$V = \pi r^2 h$

$h$ is the height
$r$ is the radius

Using these formulas, we can solve these problems.

$A_L = 2 (3.14)(48.9) (84.3) \approx \boxed{25887.856}$

$V = (3.14) (48.9)^2 (84.3) \approx \boxed{632958.069}$

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2 years ago

(Ill give brainliest if ya help me :D)

Sergio039 [100]

Answer:

A) The area of the flower bed is larger than 8 square feet

Step-by-step explanation:

This is the answer, hope it helps!

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2 years ago

A recipe for fruit salad includes 1/5 cup of grapes for 3 servings. How many cups of grapes are needed for 50 servings of this f

choli [55]

Answer:

300 is your answer

Step-by-step explanation:

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2 years ago

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

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2 years ago

Write 700 in expanded form

aleksklad [387]

Answer:

you could probably do 10x70 or (7x10) x 10 I hope this helps!

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers